I'm having trouble understand the concepts for this problem. Here's the problem:
A block is placed on a plane inclined at angle $\theta$. The coefficient of friction between the block and the plane is $\mu = \tan \theta$. The block is given a kick so that it initially moves with speed $V $horizontally along the plane (that is, in the direction perpendicular to the direction pointing straight down the plane). What is the speed of the block after a very long time?
We can see that $F_x = mg\sin \theta - mg\cos (\theta )\mu = 0$, if we gave the block an initial velocity along the plane, it should be subjected to a friction force opposite its velocity, where $f = mg \cos \theta \tan \theta = mg\sin \theta$
This however, is the answer my book gave:
The normal force from the plane is $N = mg \cos \mu$. Therefore, the friction force on the block is $N = (\tan \mu)N = mg \sin \mu$. This force acts in the direction opposite to the motion. The block also feels the gravitational force of $mg \sin \theta$ pointing down the plane. Because the magnitudes of the friction force and the gravitational force along the plane are equal, the acceleration along the direction of motion equals the negative of the acceleration in the direction down the plane. Therefore, in a small increment of time, the speed that the block loses along its direction of motion exactly equals the speed that it gains in the direction down the plane. Letting v be the speed of the block, and letting vy be the component of the velocity in the direction down the plane, we therefore have $$v + vy = C$$ where $C$ is a constant. $C$ is given by its initial value, which is $V + 0 = V$ . The final value of $C$ is $V_f + V_f = 2V_f$ (where $V_f$ is the final speed of the block), because the block is essentially moving straight down the plane after a very long time. Therefore, $2V_f = V \rightarrow V_f = \frac{V}{2}$
I don't really understand what this is trying to say. The bolded part is what's giving me problems. I think it just may be poorly written, I think I need a clearer explanation.